Dospinescu's Inequality

Question

Consider $a_1,\dots, a_n$ be positive real numbers such that $a_1 a_2 \dots a_n = 1$. Prove that $$n^n \prod_{i=1}^n \bigg ( 1 + a_i^n \bigg ) \geq \bigg( \sum_{i=1}^n a_i + \sum_{i=1}^n \frac 1 a_i \bigg )^n$$

Answer 1

By Holder $$n^n\prod_{i=1}^n\left(1+a_i^n\right)=\prod_{i=1}^n\left(n+na_i^n\right)=$$ $$=(1+1+...+1+a_1^n+1+a_1^n+a_1^n+...+a_1^n)\cdot$$ $$\cdot(1+1+...+a_2^n+1+a_2^n+1+a_2^n+...+a_2^n)\cdot...$$ $$...\cdot(a_n^n+1+...+1+1+a_n^n+a_n^n+a_n^n+...+1)\geq$$ $$\geq\left(a_n+a_{n-1}+...+a_2+a_1+\prod_{i\neq1}a_i+\prod_{i\neq2}a_i+\prod_{i\neq3}a_i...+\prod_{i\neq n}a_i\right)^n=$$ $$=\left(\sum_{i=1}^n\left(a_i+\frac{1}{a_i}\right)\right)^n$$ and we are done!